w^2+20w-7125=0

Solution for w^2+20w-7125=0 equation:

w^2+20w-7125=0

a = 1; b = 20; c = -7125;
Δ = b2-4ac
Δ = 202-4·1·(-7125)
Δ = 28900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{28900}=170$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-170}{2*1}=\frac{-190}{2}
=-95
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+170}{2*1}=\frac{150}{2}
=75
$